Transcript Circular Motion

Circular Motion
For a car going around a curve at constant
speed, the free-body diagram is:
Fcp
FN
Fw
where Fw is the weight of the car, FN is the
normal (perpendicular) force, and Fcp is the
centripetal force.
ΣFy = 0
ΣFX = Fcp
FN = Fw
 The vector sum of the forces in a vertical
direction equals zero which makes sense
because there is no vertical acceleration.
 The vector sum of the forces in a horizontal
direction is greater than zero which makes
sense because the car is accelerating in a
horizontal plane.
 Fcp is the net force which by Newton’s 2nd
law causes the car to accelerate toward the
center of the circle.
The centripetal force is given by:
Fcp = mv2/r
where m is the mass in kg, v is the velocity in
m/s, and r is the radius of the circle in m.
An object undergoing circular motion is constantly
accelerating because its instantaneous velocity is
always changing direction.
 The instantaneous velocity is always tangent
to the circle.
The centripetal acceleration is always directed
toward the center of the circle is given by:
 ac = v2/r
ac
As is the case with projectile motion, Newton’s
three laws can be found in circular motion.
Newton’s 1st law is found when the object wants
to continue its uniform motion in a straight line.
 An object does want to accelerate but can
be made to do so with a net force.
Newton’s 2nd law is found by the net force and the
acceleration acting in the same direction, toward
the center of the circle.
Newton’s 3rd law is also found but this is where a
colossal misinterpretation is made.
Remember that Newton’s 3rd law requires two
different objects.
 When you go around a corner in a car, who
or what is accelerating?
 That’s right, you and the car are accelerating
by a net force which is the centripetal force.
 Both you and the car have mass (hopefully
not the same mass) and because of inertia
both of you want a state of zero
acceleration.
 The centripetal force is the force that the
road exerts on the car (and you) via friction.
 The reaction to this force is the car exerting
a force on the road which is called the
centrifugal force.
 When the driver takes an especially sharp
turn while you are the passenger, many
people say that centrifugal force pushes
them into the door.
 This is where your “physical” intuition can
fool you.
 The centrifugal force acts on the road, not
you!
 In actuality, it is nothing more than Newton’s
1st law kicking into action.
Acceleration and Velocity
Any net force produces an acceleration which is
in the same direction as the net force.
 When the acceleration is parallel to the
velocity, the speed will increase.
 When the acceleration is in the opposite
direction of the velocity, the speed will
decrease.
 If the acceleration is perpendicular to the
velocity, the speed remains the same but
the velocity changes in direction.
Circular Motion Problem
A 0.017 kg rubber stopper is attached to a
0.89 m length of string. The stopper is swung in
a horizontal circle making one revolution in
1.2 s.
(a) Draw a free body diagram for the rubber
stopper and identify the origin of each force.
Fcp is the centripetal force and represents the
force the string exerts on the rubber stopper.
Fcp
Fw
Fw is the weight of the stopper which represents
the earth pulling down on the stopper.
Remember that in a force diagram you only
include the forces that act on the object!
(b) What is the speed of the stopper?
vave = Δx/Δt = C/Δt = 2πr/Δt
vave = 2 × 3.14 × 0.89 m/1.2 s = 4.7 m/s
(c) What is the instantaneous velocity?
v
v = 4.7 m/s tangent to the circle.
(d) What is the instantaneous acceleration?
a = v2/r = (4.7 m/s)2/0.89 m = 24 m/s2
a = 24 m/s2 toward the center of the circle.
(e) Determine both centripetal and centrifugal
force.
Fcp = ma = 0.017 kg × 24 m/s2 = 0.41 N
Fcf = Fcp = 0.41 N
(f) What happens if the string breaks?
When the string breaks, Fcp = 0, the stopper
will be in free fall and will attempt to travel in a
straight line with a velocity of 4.7 m/s while
accelerating vertically.
Vertical Circular Motion
When an object moves in a vertical circle, the
weight of an object can not be ignored.
Free-body diagram at the top of a circle:
Fw
T
Fnet = Fc
Fnet = Fw + T
Fc = Fw + T
Free-body diagram at the bottom of a circle:
T
Fw
Vertical Circle Problem
A 0.147 kg ball at the end of a 1.20 m long
string is swung in a vertical circle. If the mass
of the string is neglected:
(a) What is the minimum speed the ball must
have at the top of the circle so that it
continues to uniformly accelerate towards
the center?
m = 0.147 kg
r = 1.20 m
Fw
At the top, the minimum speed occurs when T = 0,
that is when the weight of the object supplies the
centripetal force.
Fnet = Fc = Fw
Fc = Fw
mv2 = mg
r
v = (rg)1/2 = (1.20 m × 9.80 m/s2)1/2 = 3.43 m/s
(b) What is the tension in the cord at the bottom
of its path if the ball is moving at three times
the speed at the top?
T
Fw
At the bottom, the tension is a maximum
because it must support the weight of the object
and also provide the centripetal force.
Fnet = Fc = T – Fw
Fc = T- Fw = mv2/r – mg
T = Fc + Fw
T = 0.147 kg × (10.3 m/s)2 + 0.147 kg × 9.80 m/s2
T = 17.0 N
Another Circular Motion Problem
A boy weighing 294 N sits 5.0 m from the axis
of rotation on a merry-go-round. The
merry-go-round spins at 4.7 rev/min.
(a) What is the frictional force required to keep
the boy on the merry-go-round.
Fc
FN
Fw
Fw = 294 N
Ff = Fc =
v =
r = 5.0 m
v = 4.7 rev/min
mv2
r
rev
4.7
1 min ×
2 × π × 5.0 m
1 rev
1 min
× 60 s
v = 2.5 m/s
294 N/9.80 m/s2 × (2.5 m/s)2
= 38 N
Ff =
5.0 m
(b) Determine the coefficient of friction.
µ=
Ff
FN
=
38 N
= 0.13
294 N
More Circular Motion Thoughts
In the case of a merry-go-round, friction must
provide the centripetal force.
You must always pay attention to units!
Because 1 N = kg•m/s2, all the units must be
compatible to cancel out.
One complete revolution is equal to the
circumference of the circle, 2πr.
An interesting tidbit is that the coefficient of
friction (µ) is independent of mass in this
problem.
How do you know? I’m glad you asked!
µ =
Ff
FN
=
Fc
Fw
mv2/r
= mg =
Aren’t you glad you asked?
v2
rg
Wrap Up Questions
If the average velocity of an object is zero
during a time interval, what can be said for the
displacement of that object for the same time
interval?
The displacement equals zero because
displacement is used to define velocity.
If the displacements of a particle are known at
two distinct points and the time to travel that
displacement is known, can you determine the
instantaneous velocity?
The instantaneous velocity can not be
determined unless you are told that the object is
traveling at a constant velocity, in which case
the two velocities would be the same.
You could determine the instantaneous velocity
from a Displacement vs Time graph.
If the speed of an object is constant, can it be
accelerating?
It could accelerate by following a curved path,
either going around in circular motion or turning
a corner in a car.
If the velocity of an object is constant, can it be
accelerating?
No, because acceleration is defined as the rate
at which the velocity changes.